permutation/combination problem

I'm really stuck on this problem.

It says,how many numbers between 1 and 999,999 contains each of the digit 1,2 and 3 at least once.(Hint : for each i = 1,2,3 A_i be the set of all integers from 1 to 999,999 that do not contain the digit i)

We have six digit-number* and we need to place in it digits 1, 2 and 3 in free order. We can do it in 6*5*4 ways, the remaining 3 places may include any digit, so it gives us 6*5*4*10*10*10 different numbers.

I was using the hint given in the problem. is the number of numbers from 1 to 999999 that have no 1ís.
That is : you can use any of the other nine digits, subtract 1 to account for 000000. is the number of numbers from 1 to 999999 that have no 1ís nor 2ís.
That is so forth.
The total number of numbers from 1 to 999999 is .

Try shorter way: number xxxxxx has to include each of 1, 2 and 3 digits, they can be placed in different ways:xxx321xx3x21x3xx213xxx21xxx231xx32x1x3x2x13xx2x1xx2x31xx23x1...1x32xx13x2xx1x2xx31x2x3x1x23xx132xxx12xxx312xx3x12x3xx123xxx
Each x may be replaced by any other digit, so each of 120 different cases represent 10 different numbers, that's why we get

Try shorter way: number xxxxxx has to include each of 1, 2 and 3 digits, they can be placed in different ways:Each x may be replaced by any other digit, so each of 120 different cases represent 10 different numbers, that's why we get

That happens to be a great overcount.
The correct answer is 74,460.
In the reply above, the overcount comes by allowing any other digit to be used in each case.